Question

A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds

[a]What is the expected acceleration of the car from 10,000 N force?

[b]What is the actual acceleration from the observed data of x and t?

[c]What is the difference in accelerations?

[d]What force caused this difference?

[e]What is the magnitude and direction of the force that caused the difference in acceleration?

Answer 1

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/
`s^(2)`

B. Formula: a=
`(V-V_(0) )/(t)` and s=d/t

speed= 394.6/15

s=26.3m/s

a=
`(26.3-0)/(15)`

a=1.75m/
`s^(2)`

C. 7.9-1.75=difference of 6.15m/
`s^(2)`

D. The force that most likely caused this difference is friction forces

Answer 2

Part a)

`a = 7.89 m/s^2`

Part b)

`a = 3.51 m/s^2`

Part c)

`\Deltra a = 4.38 m/s^2`

Part d)

This difference in acceleration is due to some frictional force on the surface.

Part e)

`F_f = 5552.8 N`

Step-by-step explanation:

Part a)

As we know by newton's II law

`F = ma`

here we know that

`m = 1267 kg`

`F = 10,000 N`

Now we have

`a = (F)/(m)`

`a = (10,000)/(1267)`

`a = 7.89 m/s^2`

Part b)

distance covered by the car

`d = 394.6 m`

t = 15 s

now by kinematics we have

`d = (1)/(2)at^2`

`394.6 = (1)/(2)a(15^2)`

`a = 3.51 m/s^2`

Part c)

Difference of acceleration is given as

`\Delta a = a_(expected) - a_(real)`

`\Delta a = 7.89 - 3.51`

`\Deltra a = 4.38 m/s^2`

Part d)

This difference in acceleration is due to some frictional force on the surface.

Part e)

Now for magnitude of force is given as

`F - F_f = ma`

`10,000 - F_f = ma`

`10,000 - F_f = 1267* 3.51`

`F_f = 5552.8 N`